Question

Write the partial fraction decomposition of the rational expression.

$\frac{4}{2x^2-5x-3}$

Short answer

The partial fraction decomposition of the given rational expression is given below,

$\frac{{\displaystyle 4}}{7\left(x-3\right)}-\frac{{\displaystyle 8}}{7\left(2x+1\right)}$

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\frac{4}{2x^2-5x-3}$

Step 2. Decompose the rational expression.

The denominator can be factorized as $2x^2-5x-3=\left(2x+1\right)\left(x-3\right)$.

The denominator has a repeated non-reducible quadratic factors. Then we decompose the rational expression as follows,

$\frac{4}{2x^2-5x-3}=\frac{A}{\left(x-3\right)}+\frac{B}{\left(2x+1\right)}......\left(i\right)$

Where, Aand B are to be determined.

Multiply every side by $\left(2x+1\right)\left(x-3\right)$,

$4=A\left(2x+1\right)+B\left(x-3\right)......\left(ii\right)$

Step 3. Find the value of A and B.

Assume $x=3$in equation (ii),

$$\begin{gathered} 4=A\left(7\right) \\ A=\frac{4}{7} \end{gathered}$$

Assume $x=0$in equation (ii),

role="math" localid="1647589482779" $$\begin{gathered} 4=A\left(2\left(0\right)+1\right)+B\left(\left(0\right)-3\right) \\ 4=A\left(1\right)+B\left(-3\right) \\ 4=A-3B......\left(iii\right) \end{gathered}$$

Substitute the value of Ain equation (iii),

$$\begin{gathered} 4=\frac{4}{7}-3B \\ B=-\frac{8}{7} \end{gathered}$$

Step 4. Substitute the values in equation (i).

On substituting the values of A and Bin equation (i),

$$\begin{gathered} \frac{4}{2x^2-5x-3}=\frac{{\displaystyle \frac{4}{7}}}{x-3}+\frac{{\displaystyle \frac{-8}{7}}}{2x+1} \\ \frac{4}{2x^2-5x-3}=\frac{{\displaystyle 4}}{7\left(x-3\right)}-\frac{{\displaystyle 8}}{7\left(2x+1\right)} \end{gathered}$$